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About Statistics: Complexity of statistics subject is well
known. Statistics involves solving complex problems having
multi-dimensional data using computational methods.
Information technology has played a vital role in handling and
simplifying such complex methods and scenarios but students
face a lot of difficulties in understanding the right application
of statistical concepts and implementing them using statistical
softwares. Our Statistics Assignment and Homework help
service has been strategized and simplified to help students in
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Sample Statistics Assignment and Homework Help Service Questions:
Depreciation Sample Questions
Question-1: Find the trend line equation and obtain the trend values for the following data
using the method of the least square. Also, forecast the earning for 2006.
Year :
Earning in
’000
$
1997
38
1998
40
1999
65
2000
72
2001
69
2002
60
2003
87
2004
95
Solution. Here, the number of items being 8 (i.e. even), the time deviation X will be taken
as
𝒕−𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒊𝒎𝒆
𝟏
𝟐
𝒐𝒇 𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍
to avoide the decimal numbers. Thus, the working will run as under:
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(a) Determination of the Trend Line Equation and the Trend Values
Working
The trend line equation is given by Y = a + bX
Where, a =
𝑌
𝑁
[ ∵ 𝑋𝑌 = Na + b 𝑋, and 𝑋=0]
=
526
8
= 65.75
And b =
𝑋𝑌
𝑋2 [ ∵ 𝑋𝑌 = a 𝑋 + b 𝑋2
, and 𝑋=0]
=
616
168
= 3.67 approx.
Putting the above values of a and b in the equation we get the required trend line equation
as Yc = 65.75 + 3.67 X
Where ,trend origin is 2000.5,
Y unit = annual earning, and X unit = time deviation
Putting the respective values of X in the above equation, we get the different trend values
as under:
Trend Values
For 1997 When X = -7, Yc = 65.75 + 3.67 (-7) = 40.06
Year
t
Earnings
Y
Time
dvn.
i.e.
𝒕−𝟐𝟎𝟎𝟎.𝟓
𝟏/𝟐 ×𝟏
X
XY X2
Trend
values
Yc=65.75 +
3.67X
1997
1998
1999
2000
2000.5 (mid
time)
38
40
65
72
-
69
60
87
95
-7
-5
-3
-1
0
1
3
5
7
-
266
-
200
-
195
-72
0
69
180
435
665
49
25
9
1
0
1
9
25
49
40.06
47.40
54.74
62.08
A = 65.75
69.42
76.76
84.10
91.44
Total 526 0 616 168 N = 8

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1998 When X = -5, Yc = 65.75 + 3.67 (-5) = 47.40
1999 When X =-3, Yc = 65.75 + 3.67 (-3) = 54.74
2000 When X = -1, Yc = 65.75 + 3.67 (-1) = 62.08
2001 When X =1, Yc = 65.75 + 3.67 (1) = 69.42
2002 When X = 3, Yc = 65.75 + 3.67 (3) = 76.76
2003 When X = 5, Yc = 65.75 + 3.67 (5) = 84.10
2004 When X = 7, Yc = 65.75 + 3.67 (7) = 91.44
(b) Forecasting of earnings for 2006
For 2006, X =
𝒕−𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒊𝒎𝒆
𝟏
𝟐
𝒐𝒇 𝒕𝒊𝒎𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍
=
2006−2000.5
1
2
×1
= 11
Thus, Yc = 65.75 + 3.67 (11) = 106.12
Hence, the earnings for 2005 is expected to be
= $ 106.12 × 100 = $ 106120

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Question-2: Obtain the straight line trend equation for the following data by the method of
the least square.
Year :
Sales in ’000
$
1995
140
1997
144
1998
160
1999
152
2000
168
2001
176
2004
180
Also, estimate the sales for 2002
Solution. (a) Determination of the straight line trend equation by the method of least
square
Year
t
Sales
Y
Time dvn.
i.e. t-1999 X
XY 𝑋2
1995
1997
1998
1999
2000
2001
2004
140
144
160
152
168
176
180
-4
-2
-1
0
1
2
5
-560
-288
-1
0
1
2
5
-16
4
1
0
1
4
25
Total 13994 1120 1 412 51 N = 7
Note. *In the above case, the average of the time variable is given by 𝑋 =
𝑡
𝑁
=
13994
7
=
1999 approx.
Hence, 1999 has been taken as the year of origin in the above table.
Working
The trend line equation is given by Yc = a + bX
Here, since 𝑋 ≠0, the value of the two constants a, and b are to be found out by solving
simultaneously the following two normal equations:
𝑌 = Na + b 𝑋
𝑋𝑌 = a 𝑋 + b 𝑋2
Substituting the respective values in the above we get
1120 = 7a + b
412 = a + 51b
Multiplying the eqn (ii) by 7 under the eqn (iii) and getting the same deducted from the
equation (i) we get
7a + b =1120
=
− 7𝑎+357 𝑏=2884
−356 𝑏= −1764
b =
1764
356
= 4.96

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Putting the above value of b in the equation (i) we get,
7a + 4.96 = 1120
7a = 1120 – 4.96 = 1115.04
or a = 1115.04/7 = 159.29
Putting the above values of a and b in the format of the equation we get the straight line for
trend as under:
Yc = 159.29 + 4.96X
Where, the year of working origin = 1999,
Y unit = annual sales (in ’000 $) and
X unit = time deviations.
(b)Estimation of the Sale for 2002
For 2002, X = 2002 – 1999 =3
Thus when, X = 3, Yc = 159.29 + 4.96 (3)= 159.29 + 14.88 = 174.17
Hence, the sales for 2002 are expected to be 174.17 × 103
= $ 174170.

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Question-3: Production figures of a Textile industry are as follows:
Year
Production
(in units)
:
1998
12
1999
10
2000
14
2001
11
2002
13
2003
15
2004
16
For the above data.
(i) Determine the straight line equation by change of the origin under the least square
method.
(ii) Find the trend values, and show the trend line on a graph paper, and
(iii)Estimate the production for 2005 and 2007.
Solution (i). Determination of the straight line equation by change of the origin under the
least square method.
Year
T
Prodn.
Y
Successive
values of
time
variable X
XY 𝑿 𝟐
Trend
values
T
1998
1999
2000
2001
2002
2003
2004
12
10
14
11
13
15
16
1
2
3
4
5
6
7
12
20
42
44
65
90
112
1
4
9
16
25
36
49
10.745
11.50
12.25
13.00
13.75
14.50
15.25
Total 𝑌 = 91 𝑋 =28 𝑋𝑌 = 385 𝑋2
= 140 N = 7
Note. The successive values of time variable X, have been taken as a matter of change of
the origin to reduce their magnitude for the sake of convenience in calculations.
Working
The straight line equation is given by Yc = a + bX
Here, since 𝑥 ≠ 0, we are to work out the values of the two constants, a and b by
simultaneous solution of the following two normal equations:
𝑌 = Na + b 𝑋
𝑋𝑌 = a 𝑋 + b 𝑋2
Substituting the respective values obtained from the above table in the above equation we
get,
91 = 7a + 28b
385 = 28a + 140b
Multiplying the eqn. (i) by 4 under the equation (iii) and subtracting the same from the eqn.
(ii) we get,

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28a + 140b = 385
=
− 28𝑎+112 𝑏=364
28𝑏=21
∴ b =
21
28
= .75
Putting the above value of b in the eqn. (i) we get,
7a + 28 (.75) = 91
= 7a = 91 – 21 = 70
∴ a = 70/7 = 10
Putting the above values of a and b in the relevant equation we get the straight line
equation naturalized as under:
Yc = 10 + 0.75 X
Where, X, represents successive values of the time variable, Y, the annual production and
the year of origin is 1997 the previous most year.
(ii) Calculation of the Trend values & Their Graphic Representation
1998 When X = 1, Yc = 10 + 0.75 (1) = 10.75
1999 When X =2, Yc = 10 + 0.75 (2) = 11.50
2000 When X = 3, Yc = 10 + 0.75 (3) = 12.25
2001 When X =4, Yc = 10 + 0.75 (4) = 13.00
2002 When X = 5, Yc = 10 + 0.75 (5) = 13.75
2003 When X = 6, Yc = 10 + 0.75 (6) = 14.50
2004 When X = 7, Yc = 10 + 0.75 (7)= 15.25
Graphic Representation of the Trend Line & the Original Data

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(i) Estimation the Production Figure for 2004 and 2006.
Since the last successive value of X for 2004 is 7, the successive values of X for 2005 and
2007 are 8 and 10 respectively.
Thus, for 2005, when X = 8, Yc = 10 + 0.75 (8) = 16.00
And for 2007, when X = 10, Yc = 10 + 0.75 (10) = 17.50
Hence, the estimate figures of production for 2005 and 2007 are 16000 units and 17500
units respectively.
Test of Suitability of Straight line Method
If the differences between the successive observations of a series are found to be constant,
or nearly so, the straight line model is considered to be a suitable measure for
representation of trend components, otherwise not. This fact can be determined by the
method of First Differences illustrated as under:

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Question-4. State by using the method of First Differences, if the straight line model is
suitable for finding the trend values of the following time series:
Year :
Sales :
1997
30
1998
50
1999
72
2000
90
2001
107
2002
129
2003
147
2004
170
Solution. Determination of Suitability of the Straight line model by the method of
First Differences
Year
T
Sales
Y
First Differences
1997
1998
1999
2000
2001
2002
2003
2004
30
50
72
90
107
129
147
170
50-30 = 20
72 – 50 = 22
90 – 72 = 18
107 – 90 = 17
129 – 107 = 22
147 – 129 = 18
170 – 147 = 23
From the above table, it must be seen that the first differences in the successive
observations are almost constant by 20 or nearly so. Hence, the straight line model is quite
suitable for representing the trend components of the given series.

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Vi. Parabolic Method of the least square
This method of least square is used only when the trend of a series is not linear, but
curvilinear. Under this method, a curve of parabolic type is fitted to the data to obtain their
trend values and to obtain such a curve, an equation of power series is determined in the
following model.
Yc = a + bX + cX2
+ dX3
+ ….+ mXn
It may be noted that the above equation can be carried to any power of X according to the
nature of the series. If the equation is carried only up to the second power of X, (i.e. X2
) it
is called the parabola of second degree, and if it is carried up to the 3rd power of X, (i.e.
X3) it is called the Parabola of 3rd degree. However, in actual practice the parabolic curve of
second degree is obtained in most of the cases to study the non-linear of a time series. For
this, the following equation is used.
Yc = a + bX +cX2
Where, Yc represents the computed trend value of the Y variable ‘a’ the intercept of Y, b’
the slope of the curve at the origin of X and c’, the rate of change in the slope.
In the above equation, a, b, and c are the three constants, the value of which are
determined by solving simultaneously the following three normal equations:
𝑌 = 𝑁𝑎 + 𝑏 𝑋 + 𝑐 𝑋3
It may be noted here that the first normal equation has been derived by multiplying each
set of the observed relationship by the respective coefficients of a, and getting them all
totaled ; the second normal equation has been derived by multiplying each set of the
observed relationship by the respective coefficients of b, and getting them all totaled ; and
the third equation has been derived by multiplying each set of the observed relationship by
the respective coefficients of c and getting them all totaled.
Further, it may be noted that by taking the time deviations from the midpoint of the time
variable, if 𝑋 and 𝑋3
could be made zero, the above three normal equations can be
reduced to the simplified forms to find the values of the relevant constants as follow :
From the above, it must be noticed that the value of b can be directly obtained as b =
𝑿𝒀
𝑿 𝟐,
and the values of the other two constants a and c can be obtained by solving simultaneously
the rst of the following two normal equations:
a =
𝒀− 𝒄 𝑿 𝟐
𝑿 𝟒
c =
𝑿 𝟐 𝒀−𝒂 𝑿 𝟐
𝑿 𝟒
Once, the values of the three constants a, b and c are determined in the above manner, the
trend line equation can be fitted to obtain the trend values of the given time series by
simply substituting the respective values of X therein.

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